$\begingroup$1) is true, because the normalizer condition is equivalent to every subgroup being an "infinite" version of subnormal: that is, every subgroup has an ascending series going from it to the whole group. There are definitely examples for 2), but I can't think of any. The normalizer condition (groups with it are usually called N-groups) implies local nilpotence, hence local solvability. It also implies there is an ascending series for the group with abelian factors, much like a solvable group.$\endgroup$
– user641Dec 28 '11 at 8:19

$\begingroup$@Steve: Sure that happens in nilpotent groups, and is a well-known property, but the OP should have told us which normalizer condition he meant, there could be others.$\endgroup$
– Geoff RobinsonDec 28 '11 at 8:20

$\begingroup$@Steve: Is 1. really clear? It works in nilpotent groups, but in general why couldn't you have an infinite strictly ascending chain of subgroups which never reaches the whole group?$\endgroup$
– Geoff RobinsonDec 28 '11 at 8:24

1 Answer
1

Isn't the answer to 1 clearly yes, because the normalizer of the inverse image of a subgroup of a quotient properly contains it?

As for 2, let's fix a prime $p$, and let $P_n$ be a sequence of finite $p$-groups of unbounded derived length. For example, we could take iterated wreath products. Now let $P$ be the direct product of the $P_n$. Then $P$ is not solvable, since there is no bound on its derived length.

To show that $P$ satisfies the normalizer condition, let $Q$ be a subgroup of $P$. If $Q$ does not contain $Z(P)$, which is the direct product of the $Z(P_n)$, then $H$ is properly contained in $QZ(P) \le N_P(Q)$. So $Z(P) \le Q$. But now if $P$ does not contain the second centre $Z_2(P)$ of $P$, then $Q$ is properly contained in $QZ_2(P) \le N_P(Q)$. So by induction we get $Z_r(P) \le Q$ for all $r$, but the union of the $Z_r(P)$ is $P$, so $Q=P$.